9

No, the identity map $S^2 \to S^2$ and the antipodal map are self-homotopy equivalences of $S^2$, yet the two maps are not homotopy equivalent.


5

For manifolds (or more generally absolute neighborhood retracts of euclidean spaces) consider embedding $N\subseteq \mathbb{R}^n$. Next pick a tubular neighborhood $U$ of $N$ in $\mathbb{R}^n$. There is a smooth retraction $r:U\rightarrow N$ and $U$ is an open subset of $\mathbb{R}^n$. For each $n\in \mathbb{N}$ consider the homotopy $h_n:[0,1]\times M\...


5

The situation is of course far more ridiculuous than you could possibly hope for. For instance take the paper Every finite group is the group of self-homotopy equivalences of an elliptic space, Acta Math. 213, 1, (2014), 49-62, by C. Costoya, A. Viruel. If you have not yet guessed its content: Every finite group can be realised as the group of self-homotopy ...


5

For a very simple counterexample, take $X$ and $Y$ to be discrete spaces of the same cardinality. Then any bijection between them is a homotopy equivalence (indeed, a homeomorphism), but there are no nontrivial homotopies of maps $X\to Y$.


5

The statement about complements that you claim is not true. First of all, not all topological spaces embed into $\mathbb R^n$. Even for those that do, we can find counterexamples. For instance, in $\mathbb R^2$ the subsets $A = \{(x, 0) : x \in (0, 1)\}$ and $B = \{(x, 0) : x \in \mathbb R\}$ are homeomorphic. However, $\mathbb R^2 - A$ is connected but $\...


4

In $\mathbb R$, $(0,1)$ is homeomorphic to $(1,\infty)$ (via $x \to \frac 1 x)$ but the complement of the first one is not connected whereas the complement of the second interval is connected.


3

$M$ may be constructed by taking a polygon and identifying edges in such a way that all vertices become identified with each other. If you remove a vertex from the center of the polygon, then you can retract onto the perimeter of the polygon, which is a bouquet of circles. Let $X$ denote the unit disk in $\mathbb{R}^2$ with the origin removed. Let $C$ ...


2

If $X$ is a co-H-group and $\alpha,\beta:X\rightarrow Y$ are (homotopy classes of) maps then the sum $\alpha+\beta:X\rightarrow Y$ is defined with the co-H-structure on $X$ and the induced map in homology satisfies $$(\alpha+\beta)_*=\alpha_*+\beta_*.$$ The element $r\cdot \alpha$ for $r\in\mathbb{Z}$ is obtained as the composite $$r\cdot\alpha=\alpha\circ\...


2

Your definition of $g$ is unnecessarily complicated, and this this gives rise to asking "what the intuition is for constructing such a function". It is much easier to proceed as follows. You have a homotopy $F : f \simeq c$. Let $G : c \simeq f$ be the "inverse" homotopy $G(x,t) = F(x,1-t)$. Define $$g : D^{n+1} \to Y, g(x) = \begin{cases}...


2

For a specific $t$, Try showing that the function $f:X \to I \times X$, defined by $f(x) = (t,x)$ is continuous. Then deduce that $H_t$ is continuous, using composition of functions.


2

By fibring Stasheff means Hurewicz fibration. That is a map $p:E\rightarrow B$ which has the homotopy lifting property with respect to all spaces. By equivalent to Stasheff means the following. Given any map $f:X\rightarrow Y$ there is a Hurewicz fibration $p_f:W_f\rightarrow Y$ and inverse homotopy equivalences $$X\xrightarrow{i} W_f\xrightarrow{q} X$$ such ...


2

Any good answer to this would be very long, so perhaps some people can answer with alternative perspectives. Essentially stability removes "pathologies" that only happen in low dimensions. For example, there is only one tangent bundle to a manifold, but many normal bundles. However, if we make the codimension of our embedding large enough all ...


2

Yes, it is a CW pair since $X \wedge I_+$ is the quotient of $X \times I $ by $\{*\} \times I$. It is easy to show that any point in a CW complex can be a 0-cell of a CW structure on it, so this quotient has a CW structure, and the image of $X \sqcup X$ under the quotient map is the subcomplex $X \vee X$.


2

Since you already know what the answer should be, it's maybe not so important where the operation is taking place. But here is one possible context that makes sense. As explained in the paper, there is an action of $B\mathbb{Z}$ on $\Lambda_\infty$, i.e., we have a functor $\alpha: B\mathbb{Z} \times \Lambda_\infty \to \Lambda_\infty$, which restricts to an ...


1

This is true for any two $f,g$. You do not need the fact that they are homotopic. Let $M' = M + M$ be the disjoint union of two copies of $M$. Then $f, g$ induces a smooth map $\phi : M' \to N$ by taking $f$ on the first summand and $g$ on the second. The set of regular values of $\phi$ is non-empty (it is dense in $N$). Any regular value of $\phi$ is a ...


1

Apologies, I have not fully read the question. This answer presumes that the functor lifts the normal Hom functor on both objects and morphisms. Clearly the identity element of $[X,*]$ is constant. The image of the identity of a group under a group homomorphism is the identity of the codomain. Since, $*$ is initial and the composition of any constant map ...


1

It may take a while to identify the origin of the confusion. I can no longer help with that in the space of a comment, so an answer it is. Here is the first problem that the extra image by the OP revealed. When going from picture 1 to picture 2, they seem to have in mind the homotopy shown in the animation below. Reload the page, if the animation has stopped....


1

The problem is steps $1\to 2$ and $4\to 5$. Perhaps you could imagine this problem as having a rubber band wrapped once around a metal bar (attached to a wall, say, like a ladder rung). Is there any way to get the rubber band off the bar? Or to wrap it around the bar but in the opposite direction?


1

Hint: for the right hand picture, give each $S^1$ a CW-structure of two $0$-cells on the equator, and the two $1$-cells connecting them. Take any spanning tree of the complex and contract it to a point. One idea is to take the union of bottom hemispheres of each $S^1$, and pinch them to a point.


1

Let $X$ be a topological space (if you want, you can restrict to $X = U \subset \mathbb C$). By $\mathcal P(X)$ let us denote the set of all curves $\gamma : [a,b] \to X$ which are defined on any closed interval $[a,b]$. Let $\mathcal L(X) \subset \mathcal P(X)$ denote the subset of all closed curves. We say that a closed curve $\gamma : [a,b] \to X$ is ...


1

Strictly speaking this is not true since $\emptyset$ is not contractible. So let us interpret the condition in the sense that for every real affine line $L$ in $\mathbb{R}^n$, the set $E\cap L$ is either empty or contractible. You know that $E$ is convex iff $E\cap L$ is convex for every real affine line $L$ in $\mathbb{R}^n$. Thus your first question is ...


1

I'm just going to write down a proof here, since there seem to be very few written proofs of this, since most sources seem to leave it as an exercise. Fosco of course has linked a paper on the arXiv with a proof, but I think it's best that we have an answer on MSE as well. Let $\newcommand\calC{\mathcal{C}}\calC$ be a category, let's suppose we have a closed ...


1

Yes, there is a written proof of this very long and boring exercise: https://arxiv.org/abs/1902.06074 The general statement is dubbed "Theorem 2.6", and although the obvious application is to model structures, it has absolutely nothing to do with algebraic topology, just old plain category theory.


1

As far as I can tell this is a coincidence, sort of. You can write down a homotopy that "opens" the cube by "unzipping" four edges - it would be a lot easier to draw diagrams describing this than to describe it verbally but I don't have an easy way to draw diagrams right now. In any case you don't need this. The general result is that a ...


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